I have the following integral $$g(x_0) = \int_{-\infty}^{\infty} \frac{1}{(1+x^2)^{3/4}}\frac{1}{(1+(x+x_0)^2)^{3/4}}\exp\left(-\frac{2\pi i}{\lambda}\left[\sqrt{1+x^2}-\sqrt{1+(x+x_0)^2} \right] \right)\mathrm{d}x$$

This can be re-written as $$g(x_0) =\int f(x)f^\ast(x+x_0)\mathrm{d}x$$ where $$f(x)=\frac{1}{(1+x^2)^{3/4}}\exp\left(-\frac{2\pi i}{\lambda}\sqrt{1+x^2} \right).$$

Thus, $G(\omega) = |F(\omega)|^2.$

Now, I observe from numerical integration that $g(x_0)$ is very, very, close to $$g(x_0) \approx 2\mathrm{sinc}\left( \frac{2x_0}{\lambda}\right).$$

I would like to understand why. In particular, $F(\omega)$ has a very steep decay (basically brickwall shape - resembling the sinc) around $\omega = \lambda/4$.

If the question, "why is the approximation so good?" too fuzzy, then the reason behind the super-steep decay of $F(\omega)$ can be answered.

Cheers